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CHAPTER ONE QUESTIONS AND EXERCISES

1.9 One sigma from the mean value of the Normal Distribution, from the Table(not in the book) corresponds to a value of 0.3413. This is also the probabilitythatthe random variable lies between zero and the one-sigma value, or 0.3413. This isdoubled to obtain a value from minus one sigma to plus one sigma, yielding a valueof 0.6826. One minus that value (0.3174) is the probability that the random variableis outside the minus to plus one sigma region.To move to the two sigma case:For plus two sigma, the value from the Table is 0.4772From minus two sigma to plus two sigma, the value is twice the above, or 0.9544One minus that value (0.0456) is the probability that the random variable liesoutside the minus to plus two sigma region.

To move to the three sigma case:For plus three sigma, the value from the Table is 0.49865From minus three sigma to plus three sigma, the value is twice the above, or 0.9973One minus that value (0.00270) is the probability that the random variable lies

outside the minus to plus three sigma region.To move to the four sigma case:For plus four sigma, the value from the Table is 0.499968.From minus four sigma to plus four sigma the value is twice the above, or 0.999936One minus that value (0.000064) is the probability that the random variable lies outside the minus to plus four sigma region.

1.10 sssss(x) = 2, sssss(y)=3, and s

ssss(z) = 4The basic equation for independent random variables is:

sssss2 sssss2 sssss2 sssss2

(Total) = (x) + (y) + (z)

Substituting, sssss2 (Total) = (2)2 + (3)2 + (4)2

= (4) + (9) + (16) = 29

Therefore the total error variance is 29

The square root of the variance is the standard deviation, which is 5.385



CHAPTER THREE QUESTIONS AND EXERCISES

3.4 First, calculate the cost factor (CF) and the price factor (PF)CF = (1 + FR)(1 + OH)(1 + GA) = (1.3)(1.7)(1.15) = 2.54PF = CF (1 + PR) = 2.54 (1.1) = 2.8 (approx.)We then divide the Rate per Week by 40 to obtain the Rate per Hour

Fully Loaded RateCategory of Persons Rate per Week/40 Rate per Hour(multiply by 2.8)

Senior SW Engineer 1000/40 $25 $70/hr

SW Engineer 700/40 $17.50 $49/hrDoc. Specialist 500/40 $12.50 $35/hrTraining Specialist 600/40 $15 $42/hrAverage Hourly Rate for Overall Project:

Category of Persons Rate per Hour No. Hours Total PriceSenior SW Engineer $70 400 $28,000SW Engineer $49 400 $19,600Doc. Specialist $35 120 $4,200Training Specialist $42 80 $3,360

1,000 $55,160Average Hourly Rate (fully loaded) for Entire Project = $55,160/1000 = $55.16



CHAPTER FOUR QUESTIONS AND EXERCISES

4.1 PERT ProblemCalculation of expected values for each activity:A-B: T(E) = (1+12+5)/6 == 3 weeksB-D: T(E) = (1+8+3)/6 = 2 weeksA-C: T(E) = (1+8+3)/6 = 2 weeksC-D: T(E) = (2+16+6)/6 = 4 weeksC-F: T(E) = (4+24+8)/6 = 6 weeksC-E: T(E) = (1+16+7)/6 = 4 weeksD-F: T(E) = (1+8+3)/6 = 2 weeksE-F: T(E) = (1+12+5)/6 = 3 weeksSee schedule sketch below:



4.1 (continued)Critical Path = A-C-E-FExpectedTime = 2+4+3 = 9 weeksSlack: A-B-D-F = 2 weeks Slack: A-C-F = 1 weekSlack: A-C-D-F = 1 weekStandard Deviation (on A-C-E-F)A-C: [(3-1)/6]2 = 1/9C-E: [(7-1)/6]2 = 1E-F: [(5-1)/6]2 = 4/9Overall Variance = 1/9 + 1 + 4/9 = 14/9 = 1.555Standard Deviation = vÖÖÖÖ1.555 = 1.25 weeks4.3 EVA ProblemTime Now = 18 months, Scheduled End Date = 24 months = TACBudget = $400,000 = BACSchedule Variance = $30,000 = SV = BCWP BCWSCost Variance = $20,000 = CV = BCWP ACWPBCWS = $300,000Therefore,CV = 20,000 = BCWP -ACWPSV = 30,000 = BCWP -BCWSSubtracting, CV -SV = BCWS -ACWP20,000 -30,000 = 300,000 -ACWP

and ACWP = 310,000BCWP = 20,000 + ACWP = 20,000 + 310,000 = 330,000ECAC = (ACWP)(BAC) = (310)(400,000) = $375,758(BCWP) (330)



ETAC = (BCWS)(TAC) = (300)(24) = 21.8 months(BCWP) (330)

The project is ahead in both cost and schedule (lower cost and earlier schedule) This is also indicated by the fact that the CV and SV are both positive.

In the text example, the project was behind in both cost and schedule, revealedbynegative values for CV and SV.

4.4 a. If BCWP > BCWSThe formula for SV is: SV = BCWP -BCWS, which makes SV positiveTherefore, we should be ahead of schedule

b. If ACWP > BCWPThe formula for CV is: CV = BCWP -ACWP, which makes CV negativeTherefore, we should be behind on cost, i.e., we have spent more than we haveplanned at this point in the project

c. If ACWP < BCWSThe formulas for SV and CV are:SV = BCWP -BCWS

CV = BCWP -ACWPSubtracting, we get SV -CV = ACWP BCWS, which is negativeThis implies only that the CV is greater than the SV



CHAPTER EIGHT QUESTIONS AND EXERCISES

8.1 Two additional requirements analysis steps:1.Document all cases where requirements are problematic, and reviewthese cases with appropriate internal chain of command2.Suggest alternatives to requirements found to be inadequate8.2 Functional decomposition of automobile system:1.Braking2.Motive Power3.Steering4.Environmental Control5.Safety6.Exhaust8.3 Automated Requirements Analysis Tools1. DOORS Telelogic

2. CORE Vitech Corporation8.5MTBF1 = 200 hrs, and .llll1 = 1/MTBF1 = 1/200 = .005MTBF2 = 250 hrs, and .llll2 = 1/MTBF2 = 1/250 = .004MTBF3 = 500 hrs, and .llll3 = 1/MTBF3 = 1/500 = .002MTBF4 = 1000 hrs, and .llll4 = 1/MTBF4 = 1/1000 = .001The sum of the lambdas is the overall failure rate: .005 + .004 + .002 + .001 =.012The overall MTBF is therefore 1/.012 = 83.3 hrs

8.6 LetMTBF1 = 2x, MTBF2 = 3x, MTBF3 = 4x.llll(Total) = .llll1+ .llll2+ .llll3 , and .llll= 1/MTBF.0108 = 1__ + 1__ + 1___ = 13___ 2x 3x 4x 12x12x= __13___

.0108



x = 100Therefore, MTBF1 = 2(100) = 200 hrs, .llll1 = 1/200 = .005MTBF2 = 3(100) = 300 hrs, .llll2 = 1/300 = .00333MTBF3 = 4(100) = 400 hrs, .llll3 = 1/400 = .0025Total Failure Rate = .005 + .00333 + .0025 = 0.0108

8.7 a.sssss2 (Total) <= 70, and sssssx= 4, sssssy = 6, and sssssz=?sssss2 (Total) = sssss2(x) + sssss2(y) + sssss2(z)Substituting: 70 = (4)2 + (6)2 + sssss2(z)s

ssss2(z) =70 -16 -36 =70 -52 =18sssss(z) = 4.24b.Forpart(b), T=2x +y+z2sssss22sssss2

s

ssss2 (Total) = a (x) + b2sssss2(y) +c (z)70 = (2)2(16) + 36 + sssss2(z)sssss2

(z)= 70 -64 -36 = 70 -100 = -30sssss(z) is therefore imaginary, which means there is no real number thatsatisfies the conditions of the new problem

8.8 (See also chapter one, question/exercisenumber 1.9)Plus and minus one sigma value corresponds to 68 percentPlus and minus two sigma value corresponds to 95 percentPlus and minus three sigma value corresponds to 99.7 percent



CHAPTER NINE QUESTIONS AND EXERCISES

9.6 Evaluation Criteria:a. Automobile1. Comfort and Ride2. Horsepower3. Number of Seats4. Visibility5. Fuel Consumption6. Acceleration7. Turning Radius8. Braking Performance9. Ease of Steering10. Four Wheel Drive11. Heater and Air Conditioning Performance12. Crash Performanceb. House1. Visual Appearance2. Landscaping3. Number of Rooms4. Number of Baths5. Electrical6. Plumbing

7. Maintainability8. Size of Rooms9. Garage Facility10. Number of Stairs/Floors11. Capacity of Heating and Air Conditioning12. Quality of Building Materials9.9 Confirming numbers in Table 9.8:Noise= N = 4 and sssss= 2; V= 8 andTwillvaryfrom3to7

Let D = distance = x-m (absolute value only)

For T = 3

a. D/sssss= 5/2 = 2.5 which corresponds in Gaussian Table to .4938P(d) = .5 + .4938 = .9938



b.D/sssss= 3/2P(fa) = .5For T = 4

a. D/sssssP(d) =b.D/sssssP(fa) =For T = 5

a.D/sssss= 3/2P(d) = .5 +b.D/sssss= 5/2= 1.5 which corresponds in Table to .4332

-.4332 = .0668=4/2 = 2 which corresponds in Table to .4772

.5 + .4772 = .9772= 4/2 = 2 which corresponds in Table to .4772.5 -.4772 = .0228

= 1.5 which corresponds in Table to .4332.4332 = .9332= 2.5 which corresponds in Table to .4938

P(fa) = .5 -.4938 = .0062

For T = 6

a.D/sssss= 2/2 = 1 which corresponds in Table to .3413P(d) = .5 + .3413 = .8413b.D/sssss= 6/2 = 3 which corresponds in Table to .49865P(fa) = .5 -.49865 = .00135For T = 7

a.D/sssss= ½ which corresponds in Table to .1915P(d) = .5 + .1915 = .6915b.D/sssss= 7/2 = 3.5 which corresponds in Table to .49975



P(fa) = .5 -.49975 = .00025

Confirming Numbers in Table 9.9:

V= 8, T = 5 andthermsnoise(sssss) will vary from 3.5 to 1.0

For sssss= 3.5

a.D/sssss= 3/3.5 = .857 which corresponds in Table to .303P(d) = .5 + .303 = .803b.D/sssss= 5/3.5 = 1.429 which corresponds in Table to .4235P(fa) = .5 -.4235 = .0765For sssss= 3.0

a.

D/sssss= 3/3 = 1 which corresponds in Table to .3413P(d) = .5 + .3413 = .8413b.D/sssss= 5/3 = 1.67 which corresponds in Table to .4525P(fa) = .5 -.4525 = .0475For sssss= 2.5

a.D/sssss= 3/2.5 = 1.2 which corresponds in Table to .3849

P(d) = .5 + .3849 = .8849b.D/sssss= 5/2.5 = 2 which corresponds in Table to .4772P(fa) = .5 -.4772 = .0228For sssss= 2.0

a.D/sssss= 3/2 = 1.5 which corresponds in Table to .4332P(d) = .5 + .4332 = .9332b.

D/sssss= 5/2 = 2.5 which corresponds in Table to .4938



P(fa) = .5 -.4938 = .0062

For sssss= 1.5

a.D/sssss= 3/1.5 = 2 which corresponds in Table to .4772P(d) = .5 + .4772 = .9772

b.D/sssss= 5/1.5 = 3.333 which corresponds in Table to .4995P(fa) = .5 -.4995 = .0005For sssss= 1.0

a.D/sssss= 3/1 = 3 which corresponds in Table to .49865P(d) = .5 + .49865 = .99865

b. D/s

ssss= 5/1 = 5 which corresponds in Table to .499997P(fa) = .5 -.499997 = .000003



CHAPTER TEN QUESTIONS AND EXERCISES

10.5 COCOMO ExampleGiven software project with 80,000 delivered source instructions = DSITherefore, KDSI = 80Using organic mode model:Person-months = PM = 2.4(KDSI)1.05 = 2.4(80)1.05 = 2.4(99.6) = 239 p-moTDEV = 2.5(PM)0.38 = 2.5(239)0.38 = 2.5(8.013) = 20 mos.PROD = DSI__ = 80,000 = 334.7 DSI/p-mo (vs. 346.6 for 40 KDSI)

p-mo 239FTES = p-mo__ = 239 = 11.95 persons (vs. 7.6 for 40 KDSI)TDEV 20

10.6 Software Reliability ExampleGiven initial failure intensity = 60 failures per CPU-hrAnd the slope of the Basic Execution Time Model is -0.3,And the total number of failures experienced to date is 150y = mx + bI = -.3N + 60

a.

When I = 0, N is the total number of failuresAt that point, .3N = 60 and N = 60/.3 = 200 failures or defectsb.The current failure intensity is the value of I when N = 150So we have I(now) = -.3(150) + 60 = -45 + 60 = 15 failures/CPU-hrc.The Poisson distribution equation reduces to the exponential when k = 0P(k)= (IT)k exp(-IT) = exp[-(15)(.1)] = exp (-1.5) = 0.223K!



10.7 The five scaling factors are:1. precedentedness2. development flexibility3. risk resolution4. team cohesion5. process maturityset website: sunset.usc.edu for detailed method of using these drivers to obtain numeric value of the exponent

10.8 Effort MultipliersEarly Design Effort Multipliers (7)

1. RCPX Product Reliability and Complexity2. RUSE Developed for Reusability3. PDIF Platform Difficulty4. PERS Personnel Capability5. PREX Personnel Experience6. FCIL Facilities7. SCED Required Development SchedulePost Architecture Effort Multipliers (17)

1. RELY Required Software Reliability

2. DATA Database Size3. CPLX Product Complexity4. DOCU Documentation Match to Life-Cycle Needs5. RUSE Developed for Reusability6. TIME Execution Time Constraint7. STOR Main Storage Constraint8. PVOL Platform Volatility9. ACAP Analyst Capability10. PCAP Programmer Capability11. PCON Personnel Continuity12. APEX Applications Experience13. PLEX Platform Experience14. LTEX Language and Tool Experience

15. TOOL Use of Software Tools16. SITE Multisite Development17. SCED Required Development ScheduleEffort Multipliers described in detail in: B. Boehm, et. al, Software CostEstimation with COCOMO II, 2000, Prentice Hall PTR, Upper Saddle River, N



CHAPTER ELEVEN QUESTIONS AND EXERCISES

11.1 Binomial Probability of either one or zero errorsxn-x

Binomial = P(x) =n p qx

The individual Bit Error Rate (BER) is .001

Therefore, P(0 or 1 error) =8 (.001)0(.999)8 + 8 (.001)1(.999)701

= (1)(.992) + (8)(.001)(.993)P(0 or 1 error)= .992 + (.008)(.993) = .992 + .00794 = .999944

11.2 Roulette: 18 red, 18 black, zero, double zero = 38 possibilitiesa. Prob(black) =18/38 = .4737b.

Prob(lose) = Prob (black or zero or double zero)= P(B) + P(z) + P(dz)= 18/38 + 1/38 + 1/38 = 20/38 = 0.52611.3 Normal Distribution: mean = 6, variance = 9 = sssss2, sssss=3a.Probability exceed the value 10?Distance/sigma = 4/3 = 1.33; Area (from Table lookup) = .4082Prob(x>10) = 0.5 -.4082 = .0918b.Prob(x>12)?Distance/sigma = 6/3 = 2; Area (table lookup) = .4772

Prob (x>12) = .5 -.4772 = .0228c.Prob(x>14)?



Distance/sigma = 8/3 = 2.67; Area (table lookup) = .4962Prob (x>14) = .5 -.4962 = .0038

11.4MTBF(1) = 100 hrs; .llll(1) = .01MTBF(2) = 200 hrs; .llll(2) = .005MTBF(3) = 300 hrs; .llll(3) = .00333a. Series: .llll(Total) = .llll(1) + .llll(2) + .llll(3) = .01 + .005 + .00333 = .01833P(s) = exp (-.llllt) = exp[-(.01833)(200)] = exp(-3.667) = .02556c. Parallel:P(1) = exp[-(.01)(200)] = exp(-2) = .135P(2) = exp[-(.005)(200)] = exp(-1) = .368P(3) = exp[-(.0033)(200)] = exp(-.666) = .514P(p) = 1 -(1 -.135)(1 -.368)(1 -.514) = 1 -(.865)(.632)(.486)

P(p) =1-.266 =.73411.5 .llll= .02; Repair distribution is uniform in region [2 -10]MTTR = (10+2)/2 = 6 hrsMTBF = 1/.llll= 1/.02 = 50 hrsAvailability =A = MTBF/(MTBF + MTTR) = 50/(50 + 6)A = 50/56 = .893

11.6V = 14; T = 5; N = 9,inwhichcase s

ssss=3a. For detection probability, Distance/sigma = (14 5)/3 = 9/3 = 3 whichfrom Table corresponds to a value of .49865Therefore, P(d) = .5 + .49865 = .99865



For false alarm, Distance/sigma = 5/3 = 1.67, which from TableCorresponds to a value of .4525

Therefore, P(fa) = .5 -.4525 = .0475

b. Where should threshold be for P(fa) = .02?In this case, P(fa) = .5 -Area.02 = .5 -Area, and Area = .48Going to Table yields value of Distance/sigma = D/sssss= 2.06Therefore, D = 2.06sssss= 2.06 (3) = 6.18 = Threshold in voltsGiven that threshold, Distance/sigma = D/sssss= (14-6.18)/3 = 7.82/3 = 2.61Consulting the Table at 2.61 yields an area of .49545Therefore, P(d) = .5 + .49545 = .99545

11.7Let sssss(1) = 3x; sssss(2) = 4x; sssss(3) = 5xs

ssss2sssss2sssss2sssss2(Total) = 0.5 = (1) + (2) + (3)= (3x)2 + (4x)2 + (5x)2

0.5 =9x2 + 16x2 + 25x20.5 = 50x2x2 = 0.5/50 = .01x = 0.1

Therefore, sssss(1) = 3(.1) = .3; sssss(2) = 4(.1) = .4 and sssss(3) = 5(.1) = .5

11.8 P(t) = exp(-.llllt) ; .llll= 1/MTBF, and P(t) = exp(-t/MTBF)when t = MTBF, we have P = exp(-MTBF/MTBF) = exp(-1) = .368when t = 2MTBF, we have P = exp(-2MTBF/MTBF) = exp(-2) = .135



11.9For least squares fit in text, estimate covariance of x and y, and alsocorrelation coefficient. Refer to material in text for some calculationsCov (XY) = E (XY) -E(X) E(Y)SSSSSxy = 124; SSSSSxy/n = 124/4 = 31E(X) = 17/4 = 4.25; E(Y) = 26/4 = 6.5Cov(XY) = 31 -(4.25)(6.5) = 31 -27.625Cov(XY) = 3.375Correlation Coeficient = C = Cov(XY)___ sssss(x) sssss(y)sssss2(x) = (1/4)(2-4.25)2 + (1/4)(4-4.25)2 + (1/4)(5-4.25)2 + (1/4)(6-4-25)2sssss2

(x) =8.75/4 = 2.1875 and sssss(x) = 1.479s

ssss2(y) = (1/4)(3-6.5)2 + (1/4)(6 6.5)2 + (1/4)(8 6.5)2 + (1/4)(9-6.5)2sssss2

(y) =21/4 = 5.25 and sssss(y) = 2.29C= 3.375_____ = 0.9965(1.479)(2.29)

11.10 For Rayleigh distribution, how many sigma will correspond to a

probability value of 0.5. Try also with 0.95.Rayleigh Cumulative Distribution Function (CDF) = 1 -exp(-x2/2sssss2)When P = 0.5, 0.5 = 1 -exp(-x2/2sssss2) and exp(-x2/2sssss2) = .5Using natural logs: -x 2/2sssss2 = ln (.5) = -.693x2 = 2(.693) sssss2 = 1.386 sssss2x = 1.177 sssss

When Probability = 0.95, .95 = 1 -exp(-x2/2sssss2)

exp(-x2/2sssss2) = .05



-x 2/2sssss2 = ln (.05) = -2.9957x2 = 2(2.9957) sssss2 = 5.9914 sssss2x = 2.448 sssss



CHAPTER TWELVE QUESTIONS AND EXERCISES

12.1 Categories of software tools that can be used to support (a) systems engineeringand (b) software engineeringA. Systems Engineering1. Spreadsheets2. Modeling and Simulation Tools3. Database Management Systems (DBMSs)4. Reliability Engineering5. Decision Support Tools6. Mathematical Programming7. Project Management Aids8. Statistical Analysis Tools9. Graphics Tools10. Quality Assurance ToolsB. Software Engineering1. Languages2. Structured Analysis Tools3. Requirements Analyzers4. Simulation Tools5. Reliability Estimation

6. Process/Data Flow Charting7. Various Specialized Diagramming Tools8. Software Cost Estimation9. Configuration Management Tools10. Code Generators

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